Number of times an array can be partitioned repetitively into two subarrays with equal sum

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Given an array arr[] of size N, the task is to find the number of times the array can be partitioned repetitively into two subarrays such that the sum of the elements of both the subarrays is the same.

Examples:

Input: arr[] = { 2, 2, 2, 2 }
Output: 3
Explanation:
1. Make the first partition after index 1. Remaining arrays are {2, 2} on the right side and left side both.
2. Consider the left subarray {2, 2}. Make a partition after index 0 of this left subarray.
Now two similar subarrays with one element each i.e. {2} are formed which cannot be sub-divided.
3. Consider the right subarray {2, 2}. Make a partition after index 0 of this left subarray.
Now two similar subarrays with one element each i.e. {2} are formed which cannot be sub-divided.
Hence the output is 3 as the array was partitioned 3 times.

Input: arr[] = {12, 3, 3, 0, 3, 3}
Output: 4
Explanation:
1. The first partition is after index 0. Remaining array is arr[] = {3, 3, 0, 3, 3}.
2. The second partition is after index 1. The remaining array is {3, 3}, and {0, 3, 3}.
3. The third partition is after index 0 in array {3, 3}.
4. The fourth partition is after 1 in the array {0, 3, 3}
The remaining array is {0, 3}, and {3} which cannot be sub-divided.
Hence the output is 4.

Approach: The idea is to use Recursion. Below are the steps:

Find the prefix-sum of the given array arr[] and store it in an array pref[].
Iterate from the start position to the end position.
For each possible partition index(say K), if prefix_sum[K] – prefix_sum[start-1] = prefix_sum[end] – prefix_sum[k] then the partition is valid.
If a partition is valid in the above step then proceed with the left and right sub-arrays separately and determine whether these two subarrays form a valid partition or not.
Repeat the step 3 and 4 for both the left and right partition until any further partition isn’t possible.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Recursion Function to calculate the
// possible splitting
int splitArray(int start, int end,
               int* arr,
               int* prefix_sum)
{
    // If there are less than
    // two elements, we cannot
    // partition the sub-array.
    if (start >= end)
        return 0;
 
    // Iterate from the start
    // to end-1.
    for (int k = start; k < end; ++k) {
 
        if ((prefix_sum[k] - prefix_sum[start - 1])
            == (prefix_sum[end] - prefix_sum[k])) {
 
            // Recursive call to the left
            // and the right sub-array.
            return 1 + splitArray(start,
                                  k,
                                  arr,
                                  prefix_sum)
                   + splitArray(k + 1,
                                end,
                                arr,
                                prefix_sum);
        }
    }
 
    // If there is no such partition,
    // then return 0
    return 0;
}
 
// Function to find the total splitting
void solve(int arr[], int n)
{
 
    // Prefix array to store
    // the prefix-sum using
    // 1 based indexing
    int prefix_sum[n + 1];
 
    prefix_sum[0] = 0;
 
    // Store the prefix-sum
    for (int i = 1; i <= n; ++i) {
        prefix_sum[i] = prefix_sum[i - 1]
                        + arr[i - 1];
    }
 
    // Function Call to count the
    // number of splitting
    cout << splitArray(1, n,
                       arr,
                       prefix_sum);
}
 
// Driver Code
int main()
{
    // Given array
    int arr[] = { 12, 3, 3, 0, 3, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);
 
    // Function call
    solve(arr, N);
    return 0;
}

Java

// Java program for the above approach
class GFG{
 
// Recursion Function to calculate the
// possible splitting
static int splitArray(int start, int end,
                      int[] arr,
                      int[] prefix_sum)
{
    // If there are less than
    // two elements, we cannot
    // partition the sub-array.
    if (start >= end)
        return 0;
 
    // Iterate from the start
    // to end-1.
    for (int k = start; k < end; ++k) 
    {
        if ((prefix_sum[k] - prefix_sum[start - 1]) == 
            (prefix_sum[end] - prefix_sum[k])) 
        {
 
            // Recursive call to the left
            // and the right sub-array.
            return 1 + splitArray(start, k, arr, prefix_sum) + 
                       splitArray(k + 1, end, arr, prefix_sum);
        }
    }
 
    // If there is no such partition,
    // then return 0
    return 0;
}
 
// Function to find the total splitting
static void solve(int arr[], int n)
{
 
    // Prefix array to store
    // the prefix-sum using
    // 1 based indexing
    int []prefix_sum = new int[n + 1];
 
    prefix_sum[0] = 0;
 
    // Store the prefix-sum
    for (int i = 1; i <= n; ++i)
    {
        prefix_sum[i] = prefix_sum[i - 1] + 
                               arr[i - 1];
    }
 
    // Function Call to count the
    // number of splitting
    System.out.print(splitArray(1, n, arr,
                                prefix_sum));
}
 
// Driver Code
public static void main(String[] args)
{
    // Given array
    int arr[] = { 12, 3, 3, 0, 3, 3 };
    int N = arr.length;
 
    // Function call
    solve(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

Python3

# Python3 program for the above approach
 
# Recursion Function to calculate the
# possible splitting
def splitArray(start, end, arr, prefix_sum):
     
    # If there are less than
    # two elements, we cannot
    # partition the sub-array.
    if (start >= end):
        return 0
 
    # Iterate from the start
    # to end-1.
    for k in range(start, end):
        if ((prefix_sum[k] - prefix_sum[start - 1]) ==
            (prefix_sum[end] - prefix_sum[k])) :
 
            # Recursive call to the left
            # and the right sub-array.
            return (1 + splitArray(start, k, arr,
                                   prefix_sum) +
                        splitArray(k + 1, end, arr,
                                   prefix_sum))
         
    # If there is no such partition,
    # then return 0
    return 0
 
# Function to find the total splitting
def solve(arr, n):
 
    # Prefix array to store
    # the prefix-sum using
    # 1 based indexing
    prefix_sum = [0] * (n + 1)
 
    prefix_sum[0] = 0
 
    # Store the prefix-sum
    for i in range(1, n + 1):
        prefix_sum[i] = (prefix_sum[i - 1] +
                                arr[i - 1])
     
    # Function Call to count the
    # number of splitting
    print(splitArray(1, n, arr, prefix_sum))
 
# Driver Code
 
# Given array
arr = [ 12, 3, 3, 0, 3, 3 ]
N = len(arr)
 
# Function call
solve(arr, N)
 
# This code is contributed by sanjoy_62

C#

// C# program for the above approach
using System;
 
class GFG{
 
// Recursion Function to calculate the
// possible splitting
static int splitArray(int start, int end,
                      int[] arr,
                      int[] prefix_sum)
{
     
    // If there are less than
    // two elements, we cannot
    // partition the sub-array.
    if (start >= end)
        return 0;
 
    // Iterate from the start
    // to end-1.
    for(int k = start; k < end; ++k) 
    {
       if ((prefix_sum[k] - 
            prefix_sum[start - 1]) == 
           (prefix_sum[end] - 
            prefix_sum[k])) 
       {
            
           // Recursive call to the left
           // and the right sub-array.
           return 1 + splitArray(start, k, arr, 
                                 prefix_sum) + 
                      splitArray(k + 1, end, arr,
                                 prefix_sum);
       }
    }
     
    // If there is no such partition,
    // then return 0
    return 0;
}
 
// Function to find the total splitting
static void solve(int []arr, int n)
{
     
    // Prefix array to store
    // the prefix-sum using
    // 1 based indexing
    int []prefix_sum = new int[n + 1];
 
    prefix_sum[0] = 0;
 
    // Store the prefix-sum
    for(int i = 1; i <= n; ++i)
    {
       prefix_sum[i] = prefix_sum[i - 1] + 
                              arr[i - 1];
    }
 
    // Function Call to count the
    // number of splitting
    Console.Write(splitArray(1, n, arr,
                             prefix_sum));
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array
    int []arr = { 12, 3, 3, 0, 3, 3 };
    int N = arr.Length;
 
    // Function call
    solve(arr, N);
}
}
 
// This code is contributed by Amit Katiyar

Javascript

<script>
 
// JavaScript program to implement
// the above approach
 
// Recursion Function to calculate the
// possible splitting
function splitArray(start, end, arr, prefix_sum)
{
    // If there are less than
    // two elements, we cannot
    // partition the sub-array.
    if (start >= end)
        return 0;
   
    // Iterate from the start
    // to end-1.
    for (let k = start; k < end; ++k) 
    {
        if ((prefix_sum[k] - prefix_sum[start - 1]) == 
            (prefix_sum[end] - prefix_sum[k])) 
        {
   
            // Recursive call to the left
            // and the right sub-array.
            return 1 + splitArray(start, k, arr, prefix_sum) + 
                       splitArray(k + 1, end, arr, prefix_sum);
        }
    }
   
    // If there is no such partition,
    // then return 0
    return 0;
}
   
// Function to find the total splitting
function solve(arr, n)
{
   
    // Prefix array to store
    // the prefix-sum using
    // 1 based indexing
    let prefix_sum = Array.from({length: n+1}, (_, i) => 0);
   
    prefix_sum[0] = 0;
   
    // Store the prefix-sum
    for (let i = 1; i <= n; ++i)
    {
        prefix_sum[i] = prefix_sum[i - 1] + 
                               arr[i - 1];
    }
   
    // Function Call to count the
    // number of splitting
   document.write(splitArray(1, n, arr,
                                prefix_sum));
}
 
// Driver code
 
    // Given array
    let arr = [ 12, 3, 3, 0, 3, 3 ];
    let N = arr.length;
   
    // Function call
    solve(arr, N);
 
// This code is contributed by code_hunt.
</script>

Output:

Time Complexity: O(N²)
Auxiliary Space: O(N)

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